Two identical thin metal plates has charge $q _{1}$ and $q _{2}$ respectively such that $q _{1}> q _{2}$. The plates were brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is.
JEE MAIN 2022, Medium
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Flux $\phi = EA=\frac{ q _{1}- q _{2}}{2 \varepsilon_{0}}$
Electric field between plates $E =\frac{ q _{1}- q _{2}}{2 A \varepsilon_{0}}$
$V = Ed =\frac{ q _{1}- q _{2}}{2 A \in_{0}} d$
$V =\frac{ q _{1}- q _{2}}{2 C }$
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